a: \(x^2-2x+3\)

\(=x^2-2x+1+2=\left(x-1\right)^2+2\ge2\forall x\)

=>\(A=\frac{37}{x^2-2x+3}\le\frac{37}{2}\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

b: \(x^2-5x+10\)

\(=x^2-5x+\frac{25}{4}+\frac{15}{4}\)

\(=\left(x-\frac52\right)^2+\frac{15}{4}\ge\frac{15}{4}\forall x\)

=>\(\frac{26}{x^2-5x+10}\le26:\frac{15}{4}=26\cdot\frac{4}{15}=\frac{104}{15}\forall x\)

=>\(B=-\frac{26}{x^2-5x+10}\ge-\frac{104}{15}\forall x\)

Dấu '=' xảy ra khi \(x-\frac52=0\)

=>\(x=\frac52\)

c: \(x^2-x+6\)

\(=x^2-x+\frac14+\frac{23}{4}\)

\(=\left(x-\frac12\right)^2+\frac{23}{4}\ge\frac{23}{4}\forall x\)

=>\(\frac{2023}{x^2-x+6}\le2023:\frac{23}{4}=2023\cdot\frac{4}{23}=\frac{8092}{23}\forall x\)

=>\(C=-\frac{2023}{x^2-x+6}\ge-\frac{8092}{23}\forall x\)

Dấu '=' xảy ra khi \(x-\frac12=0\)

=>\(x=\frac12\)

d: \(x^2+x+5\)

\(=x^2+x+\frac14+\frac{19}{4}\)

\(=\left(x+\frac12\right)^2+\frac{19}{4}\ge\frac{19}{4}\forall x\)

=>\(D=\frac{0.75}{x^2+x+5}\le\frac34:\frac{19}{4}=\frac{3}{19}\forall x\)

Dấu '=' xảy ra khi \(x+\frac12=0\)

=>\(x=-\frac12\)

e: \(2x^2-x+37=2\left(x^2-\frac12x+\frac{37}{2}\right)\)

\(=2\left(x^2-2\cdot x\cdot\frac14+\frac{1}{16}+\frac{295}{16}\right)=2\left(x-\frac14\right)^2+\frac{295}{8}\ge\frac{295}{8}\forall x\)

=>\(\frac{13}{2x^2-x+37}\le13:\frac{295}{8}=\frac{104}{295}\forall x\)

Dấu '=' xảy ra khi \(x-\frac14=0\)

=>\(x=\frac14\)

f: \(3x^2-x+19\)

\(=3\left(x^2-\frac13x+\frac{19}{3}\right)\)

\(=3\left(x^2-2\cdot x\cdot\frac16+\frac{1}{36}+\frac{227}{36}\right)=3\left(x-\frac16\right)^2+\frac{227}{12}\ge\frac{227}{12}\forall x\)

=>\(\frac{61}{3x^2-x+19}\le61:\frac{227}{12}=61\cdot\frac{12}{227}=\frac{732}{227}\forall x\)

=>\(-\frac{61}{3x^2-x+19}\ge-\frac{732}{227}\forall x\)

Dấu '=' xảy ra khi x-1/6=0

=>x=1/6

b: 1/2x-4=0

=>1/2x=4

hay x=8

a: x+7=0

=>x=-7

e: 4x²-81=0

=>(2x-9)(2x+9)=0

=>x=9/2 hoặc x=-9/2

g: x²-9x=0

=>x(x-9)=0

=>x=0 hoặc x=9

a)\(x+7=0=>x=-7\)

b)\(\dfrac{1}{2}x-4=0=>\dfrac{1}{2}x=4=>x=8\)

c)\(-8x+20=0=>-8x=-20=>x=\dfrac{5}{2}\)

d)\(x^2-100=0=>x^2=100=>\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)

a: x+7=0

nên x=-7

b: x-4=0

nên x=4

c: -8x+20=0

=>-8x=-20

hay x=5/2

d: x²-100=0

=>(x-10)(x+10)=0

=>x=10 hoặc x=-10

a) x +7 =0

=>x = -7

b) x - 4 =0=>x = 4

c) -8x + 20 = 0 =>-8x =-20 =>\(x=-\dfrac{20}{-8}=\dfrac{5}{2}\)

d)\(x^2-100=0=>x^2=100>\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)

$\textbf{A)}$

$A=\dfrac{4+5|1-2x|}{7}.$

Vì $|1-2x|\ge0$

$\Rightarrow A\ge\dfrac47.$

Dấu ``='' khi $1-2x=0\Leftrightarrow x=\dfrac12.$

Vậy $\min A=\dfrac47.$

(Biểu thức không có GTLN vì $|1-2x|$ có thể lớn tùy ý.)

$\textbf{B)}$

$B=\dfrac{x^2+4x-6}{3}$

$=\dfrac{(x+2)^2-10}{3}$

$\ge-\dfrac{10}{3}.$

Dấu ``='' khi $x=-2.$

Vậy $\min B=-\dfrac{10}{3}.$

(Biểu thức không có GTLN vì $(x+2)^2$ có thể lớn tùy ý.)

a: \(P\left(-1\right)=3-1+\dfrac{7}{4}=\dfrac{7}{4}+2=\dfrac{15}{4}\)

\(Q\left(\dfrac{1}{2}\right)=-3\cdot\dfrac{1}{4}+2\cdot\dfrac{1}{2}+2=-\dfrac{3}{4}+3=\dfrac{9}{4}\)

b: Đặt P(x)-Q(x)=0

\(\Leftrightarrow3x^2+x+\dfrac{7}{4}=-3x^2+2x+2\)

\(\Leftrightarrow6x^2-x-\dfrac{1}{4}=0\)

\(\Leftrightarrow24x^2-4x-1=0\)

\(\text{Δ}=\left(-4\right)^2-4\cdot24\cdot\left(-1\right)=112>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{4-4\sqrt{7}}{48}=\dfrac{1-\sqrt{7}}{12}\\x_2=\dfrac{1+\sqrt{7}}{12}\end{matrix}\right.\)

a) Đặt A(x)=0

\(\Leftrightarrow4x-4+3x-5=0\)

\(\Leftrightarrow7x=9\)

hay \(x=\dfrac{9}{7}\)

b) Đặt B(x)=0

\(\Leftrightarrow-1\dfrac{1}{3}x^2+x=0\)

\(\Leftrightarrow x\left(-\dfrac{4}{3}x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-\dfrac{4}{3}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\)

Dùng sai mục đích dấu nha em, mình phải dùng ngoặc vuông chứ không phải nhọn nha!

a: \(\frac{x}{x-y}=\frac{x\cdot\left(x-y\right)^2}{\left(x-y\right)\left(x-y\right)^2}=\frac{x\left(x-y\right)^2}{\left(x-y\right)^3}\)

\(\frac{y}{\left(x-y\right)^2}=\frac{y\left(x-y\right)}{\left(x-y\right)^2\cdot\left(x-y\right)}=\frac{y\left(x-y\right)}{\left(x-y\right)^3}\)

\(\frac{1}{\left(y-x\right)^3}=\frac{-1}{\left(x-y\right)^3}\)

b: \(\frac{1}{2x+4}=\frac{1}{2\left(x+2\right)}=\frac{1\cdot\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}=\frac{x-2}{2\left(x+2\right)\left(x-2\right)}\)

\(\frac{x}{2x-4}=\frac{x}{2\left(x-2\right)}=\frac{x\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}\)

\(\frac{3}{4-x^2}=\frac{-3}{x^2-4}=\frac{-3}{\left(x-2\right)\left(x+2\right)}=\frac{-3\cdot2}{2\left(x-2\right)\left(x+2\right)}=\frac{-6}{2\left(x-2\right)\left(x+2\right)}\)

a: \(\frac{x}{x-y}=\frac{x\cdot\left(x-y\right)^2}{\left(x-y\right)\left(x-y\right)^2}=\frac{x\left(x-y\right)^2}{\left(x-y\right)^3}\)

\(\frac{y}{\left(x-y\right)^2}=\frac{y\left(x-y\right)}{\left(x-y\right)^2\cdot\left(x-y\right)}=\frac{y\left(x-y\right)}{\left(x-y\right)^3}\)

\(\frac{1}{\left(y-x\right)^3}=\frac{-1}{\left(x-y\right)^3}\)

b: \(\frac{1}{2x+4}=\frac{1}{2\left(x+2\right)}=\frac{1\cdot\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}=\frac{x-2}{2\left(x+2\right)\left(x-2\right)}\)

\(\frac{x}{2x-4}=\frac{x}{2\left(x-2\right)}=\frac{x\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}\)

\(\frac{3}{4-x^2}=\frac{-3}{x^2-4}=\frac{-3}{\left(x-2\right)\left(x+2\right)}=\frac{-3\cdot2}{2\left(x-2\right)\left(x+2\right)}=\frac{-6}{2\left(x-2\right)\left(x+2\right)}\)

Ta có:
\(4x^2+\dfrac{2}{5}x\)
\(=x\left(4x+\dfrac{2}{5}\right)\)
Do đó để đa thức \(4x^2+\dfrac{2}{5}x\) có nghiệm thì \(x\left(4x+\dfrac{2}{5}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\4x+\dfrac{2}{5}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\4x=-\dfrac{2}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{10}\end{matrix}\right.\)
Vậy nghiệm của đa thức là \(x\in\left\{0;-\dfrac{1}{10}\right\}\)

=>2x(2x+1/5)=0

=>x=0 hoặc x=-1/10